Integrand size = 29, antiderivative size = 157 \[ \int \frac {\cos ^4(e+f x) \sin ^3(e+f x)}{(a+a \sin (e+f x))^8} \, dx=\frac {4 \cos (e+f x)}{11 a^8 f (1+\sin (e+f x))^6}-\frac {52 \cos (e+f x)}{33 a^8 f (1+\sin (e+f x))^5}+\frac {617 \cos (e+f x)}{231 a^8 f (1+\sin (e+f x))^4}-\frac {846 \cos (e+f x)}{385 a^8 f (1+\sin (e+f x))^3}+\frac {1003 \cos (e+f x)}{1155 a^8 f (1+\sin (e+f x))^2}-\frac {152 \cos (e+f x)}{1155 a^8 f (1+\sin (e+f x))} \] Output:
4/11*cos(f*x+e)/a^8/f/(1+sin(f*x+e))^6-52/33*cos(f*x+e)/a^8/f/(1+sin(f*x+e ))^5+617/231*cos(f*x+e)/a^8/f/(1+sin(f*x+e))^4-846/385*cos(f*x+e)/a^8/f/(1 +sin(f*x+e))^3+1003/1155*cos(f*x+e)/a^8/f/(1+sin(f*x+e))^2-152/1155*cos(f* x+e)/a^8/f/(1+sin(f*x+e))
Time = 3.96 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.24 \[ \int \frac {\cos ^4(e+f x) \sin ^3(e+f x)}{(a+a \sin (e+f x))^8} \, dx=-\frac {-486024 \cos \left (e+\frac {f x}{2}\right )+351450 \cos \left (e+\frac {3 f x}{2}\right )+180015 \cos \left (3 e+\frac {5 f x}{2}\right )-63580 \cos \left (3 e+\frac {7 f x}{2}\right )-15004 \cos \left (5 e+\frac {9 f x}{2}\right )+1975 \cos \left (5 e+\frac {11 f x}{2}\right )-425964 \sin \left (\frac {f x}{2}\right )-299970 \sin \left (2 e+\frac {3 f x}{2}\right )+145695 \sin \left (2 e+\frac {5 f x}{2}\right )+44990 \sin \left (4 e+\frac {7 f x}{2}\right )-6710 \sin \left (4 e+\frac {9 f x}{2}\right )+\sin \left (6 e+\frac {11 f x}{2}\right )}{240240 a^8 f \left (\cos \left (\frac {e}{2}\right )+\sin \left (\frac {e}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^{11}} \] Input:
Integrate[(Cos[e + f*x]^4*Sin[e + f*x]^3)/(a + a*Sin[e + f*x])^8,x]
Output:
-1/240240*(-486024*Cos[e + (f*x)/2] + 351450*Cos[e + (3*f*x)/2] + 180015*C os[3*e + (5*f*x)/2] - 63580*Cos[3*e + (7*f*x)/2] - 15004*Cos[5*e + (9*f*x) /2] + 1975*Cos[5*e + (11*f*x)/2] - 425964*Sin[(f*x)/2] - 299970*Sin[2*e + (3*f*x)/2] + 145695*Sin[2*e + (5*f*x)/2] + 44990*Sin[4*e + (7*f*x)/2] - 67 10*Sin[4*e + (9*f*x)/2] + Sin[6*e + (11*f*x)/2])/(a^8*f*(Cos[e/2] + Sin[e/ 2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^11)
Time = 0.81 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {3042, 3354, 3042, 3351, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\sin ^3(e+f x) \cos ^4(e+f x)}{(a \sin (e+f x)+a)^8} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (e+f x)^3 \cos (e+f x)^4}{(a \sin (e+f x)+a)^8}dx\) |
| \(\Big \downarrow \) 3354 |
| \(\displaystyle \frac {\int \sec ^9(e+f x) (a-a \sin (e+f x))^8 \tan ^3(e+f x)dx}{a^{16}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\sin (e+f x)^3 (a-a \sin (e+f x))^8}{\cos (e+f x)^{12}}dx}{a^{16}}\) |
| \(\Big \downarrow \) 3351 |
| \(\displaystyle \frac {\int \left (\frac {1}{a^4 (\sin (e+f x)+1)}-\frac {7}{a^4 (\sin (e+f x)+1)^2}+\frac {19}{a^4 (\sin (e+f x)+1)^3}-\frac {25}{a^4 (\sin (e+f x)+1)^4}+\frac {16}{a^4 (\sin (e+f x)+1)^5}-\frac {4}{a^4 (\sin (e+f x)+1)^6}\right )dx}{a^4}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {152 \cos (e+f x)}{1155 a^4 f (\sin (e+f x)+1)}+\frac {1003 \cos (e+f x)}{1155 a^4 f (\sin (e+f x)+1)^2}-\frac {846 \cos (e+f x)}{385 a^4 f (\sin (e+f x)+1)^3}+\frac {617 \cos (e+f x)}{231 a^4 f (\sin (e+f x)+1)^4}-\frac {52 \cos (e+f x)}{33 a^4 f (\sin (e+f x)+1)^5}+\frac {4 \cos (e+f x)}{11 a^4 f (\sin (e+f x)+1)^6}}{a^4}\) |
Input:
Int[(Cos[e + f*x]^4*Sin[e + f*x]^3)/(a + a*Sin[e + f*x])^8,x]
Output:
((4*Cos[e + f*x])/(11*a^4*f*(1 + Sin[e + f*x])^6) - (52*Cos[e + f*x])/(33* a^4*f*(1 + Sin[e + f*x])^5) + (617*Cos[e + f*x])/(231*a^4*f*(1 + Sin[e + f *x])^4) - (846*Cos[e + f*x])/(385*a^4*f*(1 + Sin[e + f*x])^3) + (1003*Cos[ e + f*x])/(1155*a^4*f*(1 + Sin[e + f*x])^2) - (152*Cos[e + f*x])/(1155*a^4 *f*(1 + Sin[e + f*x])))/a^4
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[1/a^p Int[Expan dTrig[(d*sin[e + f*x])^n*(a - b*sin[e + f*x])^(p/2)*(a + b*sin[e + f*x])^(m + p/2), x], x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && In tegersQ[m, n, p/2] && ((GtQ[m, 0] && GtQ[p, 0] && LtQ[-m - p, n, -1]) || (G tQ[m, 2] && LtQ[p, 0] && GtQ[m + p/2, 0]))
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n _)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(a/g)^(2* m) Int[(g*Cos[e + f*x])^(2*m + p)*((d*Sin[e + f*x])^n/(a - b*Sin[e + f*x] )^m), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && ILtQ[m, 0]
Time = 1.26 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.72
| method | result | size |
| parallelrisch | \(\frac {-\frac {4}{1155}-\frac {4 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{105}-\frac {4 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{21}-\frac {4 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{7}+\frac {20 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{7}-\frac {44 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{5}+\frac {36 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{5}-4 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{f \,a^{8} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{11}}\) | \(113\) |
| derivativedivides | \(\frac {\frac {256}{11 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{11}}-\frac {128}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{10}}+\frac {2064}{7 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{7}}+\frac {176}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}-\frac {384}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{8}}-\frac {136}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{6}}+\frac {896}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{9}}-\frac {4}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}}{f \,a^{8}}\) | \(130\) |
| default | \(\frac {\frac {256}{11 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{11}}-\frac {128}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{10}}+\frac {2064}{7 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{7}}+\frac {176}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}-\frac {384}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{8}}-\frac {136}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{6}}+\frac {896}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{9}}-\frac {4}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}}{f \,a^{8}}\) | \(130\) |
| risch | \(-\frac {2 \left (3465 i {\mathrm e}^{9 i \left (f x +e \right )}+1155 \,{\mathrm e}^{10 i \left (f x +e \right )}-23100 i {\mathrm e}^{7 i \left (f x +e \right )}-13860 \,{\mathrm e}^{8 i \left (f x +e \right )}+32802 i {\mathrm e}^{5 i \left (f x +e \right )}+37422 \,{\mathrm e}^{6 i \left (f x +e \right )}-11220 i {\mathrm e}^{3 i \left (f x +e \right )}-27060 \,{\mathrm e}^{4 i \left (f x +e \right )}+517 i {\mathrm e}^{i \left (f x +e \right )}+4895 \,{\mathrm e}^{2 i \left (f x +e \right )}-152\right )}{1155 f \,a^{8} \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )^{11}}\) | \(140\) |
Input:
int(cos(f*x+e)^4*sin(f*x+e)^3/(a+a*sin(f*x+e))^8,x,method=_RETURNVERBOSE)
Output:
4/1155*(-1-11*tan(1/2*f*x+1/2*e)-55*tan(1/2*f*x+1/2*e)^2-165*tan(1/2*f*x+1 /2*e)^3+825*tan(1/2*f*x+1/2*e)^4-2541*tan(1/2*f*x+1/2*e)^5+2079*tan(1/2*f* x+1/2*e)^6-1155*tan(1/2*f*x+1/2*e)^7)/f/a^8/(tan(1/2*f*x+1/2*e)+1)^11
Leaf count of result is larger than twice the leaf count of optimal. 291 vs. \(2 (145) = 290\).
Time = 0.07 (sec) , antiderivative size = 291, normalized size of antiderivative = 1.85 \[ \int \frac {\cos ^4(e+f x) \sin ^3(e+f x)}{(a+a \sin (e+f x))^8} \, dx=\frac {152 \, \cos \left (f x + e\right )^{6} - 243 \, \cos \left (f x + e\right )^{5} - 745 \, \cos \left (f x + e\right )^{4} + 455 \, \cos \left (f x + e\right )^{3} + 1015 \, \cos \left (f x + e\right )^{2} + {\left (152 \, \cos \left (f x + e\right )^{5} + 395 \, \cos \left (f x + e\right )^{4} - 350 \, \cos \left (f x + e\right )^{3} - 805 \, \cos \left (f x + e\right )^{2} + 210 \, \cos \left (f x + e\right ) + 420\right )} \sin \left (f x + e\right ) - 210 \, \cos \left (f x + e\right ) - 420}{1155 \, {\left (a^{8} f \cos \left (f x + e\right )^{6} - 5 \, a^{8} f \cos \left (f x + e\right )^{5} - 18 \, a^{8} f \cos \left (f x + e\right )^{4} + 20 \, a^{8} f \cos \left (f x + e\right )^{3} + 48 \, a^{8} f \cos \left (f x + e\right )^{2} - 16 \, a^{8} f \cos \left (f x + e\right ) - 32 \, a^{8} f - {\left (a^{8} f \cos \left (f x + e\right )^{5} + 6 \, a^{8} f \cos \left (f x + e\right )^{4} - 12 \, a^{8} f \cos \left (f x + e\right )^{3} - 32 \, a^{8} f \cos \left (f x + e\right )^{2} + 16 \, a^{8} f \cos \left (f x + e\right ) + 32 \, a^{8} f\right )} \sin \left (f x + e\right )\right )}} \] Input:
integrate(cos(f*x+e)^4*sin(f*x+e)^3/(a+a*sin(f*x+e))^8,x, algorithm="frica s")
Output:
1/1155*(152*cos(f*x + e)^6 - 243*cos(f*x + e)^5 - 745*cos(f*x + e)^4 + 455 *cos(f*x + e)^3 + 1015*cos(f*x + e)^2 + (152*cos(f*x + e)^5 + 395*cos(f*x + e)^4 - 350*cos(f*x + e)^3 - 805*cos(f*x + e)^2 + 210*cos(f*x + e) + 420) *sin(f*x + e) - 210*cos(f*x + e) - 420)/(a^8*f*cos(f*x + e)^6 - 5*a^8*f*co s(f*x + e)^5 - 18*a^8*f*cos(f*x + e)^4 + 20*a^8*f*cos(f*x + e)^3 + 48*a^8* f*cos(f*x + e)^2 - 16*a^8*f*cos(f*x + e) - 32*a^8*f - (a^8*f*cos(f*x + e)^ 5 + 6*a^8*f*cos(f*x + e)^4 - 12*a^8*f*cos(f*x + e)^3 - 32*a^8*f*cos(f*x + e)^2 + 16*a^8*f*cos(f*x + e) + 32*a^8*f)*sin(f*x + e))
Timed out. \[ \int \frac {\cos ^4(e+f x) \sin ^3(e+f x)}{(a+a \sin (e+f x))^8} \, dx=\text {Timed out} \] Input:
integrate(cos(f*x+e)**4*sin(f*x+e)**3/(a+a*sin(f*x+e))**8,x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 401 vs. \(2 (145) = 290\).
Time = 0.05 (sec) , antiderivative size = 401, normalized size of antiderivative = 2.55 \[ \int \frac {\cos ^4(e+f x) \sin ^3(e+f x)}{(a+a \sin (e+f x))^8} \, dx=-\frac {4 \, {\left (\frac {11 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {55 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {165 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - \frac {825 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {2541 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} - \frac {2079 \, \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} + \frac {1155 \, \sin \left (f x + e\right )^{7}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{7}} + 1\right )}}{1155 \, {\left (a^{8} + \frac {11 \, a^{8} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {55 \, a^{8} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {165 \, a^{8} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {330 \, a^{8} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {462 \, a^{8} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + \frac {462 \, a^{8} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} + \frac {330 \, a^{8} \sin \left (f x + e\right )^{7}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{7}} + \frac {165 \, a^{8} \sin \left (f x + e\right )^{8}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{8}} + \frac {55 \, a^{8} \sin \left (f x + e\right )^{9}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{9}} + \frac {11 \, a^{8} \sin \left (f x + e\right )^{10}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{10}} + \frac {a^{8} \sin \left (f x + e\right )^{11}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{11}}\right )} f} \] Input:
integrate(cos(f*x+e)^4*sin(f*x+e)^3/(a+a*sin(f*x+e))^8,x, algorithm="maxim a")
Output:
-4/1155*(11*sin(f*x + e)/(cos(f*x + e) + 1) + 55*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 165*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 825*sin(f*x + e)^4/ (cos(f*x + e) + 1)^4 + 2541*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 - 2079*sin (f*x + e)^6/(cos(f*x + e) + 1)^6 + 1155*sin(f*x + e)^7/(cos(f*x + e) + 1)^ 7 + 1)/((a^8 + 11*a^8*sin(f*x + e)/(cos(f*x + e) + 1) + 55*a^8*sin(f*x + e )^2/(cos(f*x + e) + 1)^2 + 165*a^8*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3 30*a^8*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 462*a^8*sin(f*x + e)^5/(cos(f *x + e) + 1)^5 + 462*a^8*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 330*a^8*sin (f*x + e)^7/(cos(f*x + e) + 1)^7 + 165*a^8*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 + 55*a^8*sin(f*x + e)^9/(cos(f*x + e) + 1)^9 + 11*a^8*sin(f*x + e)^10 /(cos(f*x + e) + 1)^10 + a^8*sin(f*x + e)^11/(cos(f*x + e) + 1)^11)*f)
Time = 0.23 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.71 \[ \int \frac {\cos ^4(e+f x) \sin ^3(e+f x)}{(a+a \sin (e+f x))^8} \, dx=-\frac {4 \, {\left (1155 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{7} - 2079 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} + 2541 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 825 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 165 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 55 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 11 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}}{1155 \, a^{8} f {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{11}} \] Input:
integrate(cos(f*x+e)^4*sin(f*x+e)^3/(a+a*sin(f*x+e))^8,x, algorithm="giac" )
Output:
-4/1155*(1155*tan(1/2*f*x + 1/2*e)^7 - 2079*tan(1/2*f*x + 1/2*e)^6 + 2541* tan(1/2*f*x + 1/2*e)^5 - 825*tan(1/2*f*x + 1/2*e)^4 + 165*tan(1/2*f*x + 1/ 2*e)^3 + 55*tan(1/2*f*x + 1/2*e)^2 + 11*tan(1/2*f*x + 1/2*e) + 1)/(a^8*f*( tan(1/2*f*x + 1/2*e) + 1)^11)
Time = 18.08 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.31 \[ \int \frac {\cos ^4(e+f x) \sin ^3(e+f x)}{(a+a \sin (e+f x))^8} \, dx=-\frac {4\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left ({\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7+11\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )+55\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+165\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3-825\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+2541\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5-2079\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6+1155\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7\right )}{1155\,a^8\,f\,{\left (\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )+\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}^{11}} \] Input:
int((cos(e + f*x)^4*sin(e + f*x)^3)/(a + a*sin(e + f*x))^8,x)
Output:
-(4*cos(e/2 + (f*x)/2)^4*(cos(e/2 + (f*x)/2)^7 + 1155*sin(e/2 + (f*x)/2)^7 - 2079*cos(e/2 + (f*x)/2)*sin(e/2 + (f*x)/2)^6 + 11*cos(e/2 + (f*x)/2)^6* sin(e/2 + (f*x)/2) + 2541*cos(e/2 + (f*x)/2)^2*sin(e/2 + (f*x)/2)^5 - 825* cos(e/2 + (f*x)/2)^3*sin(e/2 + (f*x)/2)^4 + 165*cos(e/2 + (f*x)/2)^4*sin(e /2 + (f*x)/2)^3 + 55*cos(e/2 + (f*x)/2)^5*sin(e/2 + (f*x)/2)^2))/(1155*a^8 *f*(cos(e/2 + (f*x)/2) + sin(e/2 + (f*x)/2))^11)
Time = 0.17 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.54 \[ \int \frac {\cos ^4(e+f x) \sin ^3(e+f x)}{(a+a \sin (e+f x))^8} \, dx=\frac {-4 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}+\frac {36 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{5}-\frac {44 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{5}+\frac {20 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{7}-\frac {4 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{7}-\frac {4 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{21}-\frac {4 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{105}-\frac {4}{1155}}{a^{8} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{11}+11 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{10}+55 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}+165 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}+330 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}+462 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}+462 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}+330 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+165 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}+55 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+11 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )} \] Input:
int(cos(f*x+e)^4*sin(f*x+e)^3/(a+a*sin(f*x+e))^8,x)
Output:
(4*( - 1155*tan((e + f*x)/2)**7 + 2079*tan((e + f*x)/2)**6 - 2541*tan((e + f*x)/2)**5 + 825*tan((e + f*x)/2)**4 - 165*tan((e + f*x)/2)**3 - 55*tan(( e + f*x)/2)**2 - 11*tan((e + f*x)/2) - 1))/(1155*a**8*f*(tan((e + f*x)/2)* *11 + 11*tan((e + f*x)/2)**10 + 55*tan((e + f*x)/2)**9 + 165*tan((e + f*x) /2)**8 + 330*tan((e + f*x)/2)**7 + 462*tan((e + f*x)/2)**6 + 462*tan((e + f*x)/2)**5 + 330*tan((e + f*x)/2)**4 + 165*tan((e + f*x)/2)**3 + 55*tan((e + f*x)/2)**2 + 11*tan((e + f*x)/2) + 1))